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timama [110]
2 years ago
12

What is m<4 if m<5 = (2x) and m<4 = (x + 9)º? Please

Mathematics
2 answers:
dolphi86 [110]2 years ago
8 0

m∠4= 2x°

m∠5 = (x+9)°

case one: If both lie on the same plane then,

m∠4+ m∠5 = 180°

2x+x +9° = 180°

3x+9° = 180°

3x=171°

x= 171°/3

x= 57°

so,

m∠4 = 57° x 2° = 114°

m∠5= 57°+9° =66°

case two : If the angles are on opposite side of each other then,

m∠4=m∠5

2x= x+9°

2x-x=9°

x=9°

so,

m∠4m= 2x9 = 18°

∠5= 9+9= 18°

Maslowich2 years ago
6 0

1.) m∠4 = 2x°

→m∠4+ m∠5 = 180°

→2x+x +9° = 180°

→3x+9° = 180°

→3x=171°

→x= \frac{171°}{3}

→x= 57°

So,

m∠4 = 57° × 2° = 114°

m∠5= 57°+9° =66°

2.) m∠5 = (x+9)°

→m∠4=m∠5

→2x= x+9°

→2x-x=9°

→x=9°

So,

m∠4m= 2×9 = 18°

∠5= 9+9= 18°

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among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
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Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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